R 使用列存根字符串重塑宽到长
R Reshape Wide To Long Using Column Stub Strings
data1=data.frame("School"=c(1,1,2,2,3,3,4,4),
"Fund"=c(0,1,0,1,0,1,0,1),
"Total_A_Grade5"=c(22,20,21,24,24,26,25,22),
"Group1_A_Grade5"=c(10,6,6,10,9,9,9,10),
"Group2_A_Grade5"=c(5,9,9,8,10,8,8,6),
"Total_B_Grade5"=c(23,33,19,21,19,23,20,21),
"Group1_B_Grade5"=c(8,7,7,10,9,9,5,5),
"Group2_B_Grade5"=c(6,10,7,6,6,5,9,9),
"Total_A_Grade6"=c(18,24,16,24,26,25,16,19),
"Group1_A_Grade6"=c(7,7,5,9,10,9,5,7),
"Group2_A_Grade6"=c(5,8,6,7,10,8,8,9),
"Total_B_Grade6"=c(26,23,22,24,21,22,24,19),
"Group1_B_Grade6"=c(10,10,6,10,7,8,8,7),
"Group2_B_Grade6"=c(9,6,9,6,7,6,9,9),
"Total_A_Grade7"=c(20,19,18,25,16,21,19,26),
"Group1_A_Grade7"=c(9,7,7,9,7,7,5,8),
"Group2_A_Grade7"=c(8,5,7,9,6,5,5,9),
"Total_B_Grade7"=c(25,21,24,25,18,18,27,18),
"Group1_B_Grade7"=c(10,10,10,7,5,6,8,5),
"Group2_B_Grade7"=c(9,6,8,10,8,6,10,6))
data2=data.frame("School"=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1),
"Fund"=c(0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1),
"Type"=c('Total','Total','Group1','Group1','Group2','Group2','Total','Total','Group1','Group1','Group2','Group2','Total','Total','Group1','Group1','Group2','Group2','Total','Total','Group1','Group1','Group2','Group2'),
"Class"=c('A','A','A','A','A','A','B','B','B','B','B','B','A','A','A','A','A','A','B','B','B','B','B','B'),
"Grade"=c(5,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6),
"Score"=c(22,20,10,6,5,9,23,33,8,7,6,10,18,24,7,7,5,8,26,23,10,10,9,6))
我有 'data1' 并且想要重塑以制作 'data2'，它只是展示了 School 1 5 年级和 6 年级的示例，但我想要重塑所有 data1。
'data1'的列名包含丰富的信息。例如，Group2_B_Grade6 表示 'Type' = Group2，'Class' = B，'Grade' = 6。我希望重塑 'data1'，然后使用这些由“分隔的存根” _" 作为 colnames 准备 'data2'
data3=data.frame("School"=c(1,1,2,2,3,3,4,4),
"Fund"=c(0,1,0,1,0,1,0,1),
"Grade_5"=c(22,20,21,24,24,26,25,22),
"Grade_6"=c(10,6,6,10,9,9,9,10),
"Grade_7"=c(5,9,9,8,10,8,8,6))
使用dplyr（和tidyr）：
library(dplyr)
library(tidyr)
data2 <- data1 %>%
pivot_longer(-c(School, Fund)) %>%
separate(name, into = c('Type', 'Class', 'Grade')) %>%
extract(Grade, 'Grade', "([0-9]+)")
data2
#> # A tibble: 144 x 6
#> School Fund Type Class Grade value
#> <dbl> <dbl> <chr> <chr> <chr> <dbl>
#> 1 1 0 Total A 5 22
#> 2 1 0 Group1 A 5 10
#> 3 1 0 Group2 A 5 5
#> 4 1 0 Total B 5 23
#> 5 1 0 Group1 B 5 8
#> 6 1 0 Group2 B 5 6
#> 7 1 0 Total A 6 18
#> 8 1 0 Group1 A 6 7
#> 9 1 0 Group2 A 6 5
#> 10 1 0 Total B 6 26
#> # … with 134 more rows
由 reprex package (v0.3.0)
于 2020-04-06 创建
您可以直接使用 pivot_longer 和 names_pattern 中的一些正则表达式来执行此操作。
tidyr::pivot_longer(data1,
cols = -c(School, Fund),
names_to = c('Type', 'Class', 'Grade'),
names_pattern = '(.*?)_([A-Z])_Grade(\d+)',
values_to = 'Score')
# A tibble: 144 x 6
# School Fund Type Class Grade Score
# <dbl> <dbl> <chr> <chr> <chr> <dbl>
# 1 1 0 Total A 5 22
# 2 1 0 Group1 A 5 10
# 3 1 0 Group2 A 5 5
# 4 1 0 Total B 5 23
# 5 1 0 Group1 B 5 8
# 6 1 0 Group2 B 5 6
# 7 1 0 Total A 6 18
# 8 1 0 Group1 A 6 7
# 9 1 0 Group2 A 6 5
#10 1 0 Total B 6 26
# … with 134 more rows
我们可以使用 melt 来自 data.table
library(data.table)
melt(setDT(data1), id.var = c('School', 'Fund'))[,
c('Type', 'Class', 'Grade') := tstrsplit(variable, "_")][,
Grade := sub('Grade', '', Grade)][, variable := NULL][]
# School Fund value Type Class Grade
# 1: 1 0 22 Total A 5
# 2: 1 1 20 Total A 5
# 3: 2 0 21 Total A 5
# 4: 2 1 24 Total A 5
# 5: 3 0 24 Total A 5
# ---
#140: 2 1 10 Group2 B 7
#141: 3 0 8 Group2 B 7
#142: 3 1 6 Group2 B 7
#143: 4 0 10 Group2 B 7
#144: 4 1 6 Group2 B 7
data1=data.frame("School"=c(1,1,2,2,3,3,4,4),
"Fund"=c(0,1,0,1,0,1,0,1),
"Total_A_Grade5"=c(22,20,21,24,24,26,25,22),
"Group1_A_Grade5"=c(10,6,6,10,9,9,9,10),
"Group2_A_Grade5"=c(5,9,9,8,10,8,8,6),
"Total_B_Grade5"=c(23,33,19,21,19,23,20,21),
"Group1_B_Grade5"=c(8,7,7,10,9,9,5,5),
"Group2_B_Grade5"=c(6,10,7,6,6,5,9,9),
"Total_A_Grade6"=c(18,24,16,24,26,25,16,19),
"Group1_A_Grade6"=c(7,7,5,9,10,9,5,7),
"Group2_A_Grade6"=c(5,8,6,7,10,8,8,9),
"Total_B_Grade6"=c(26,23,22,24,21,22,24,19),
"Group1_B_Grade6"=c(10,10,6,10,7,8,8,7),
"Group2_B_Grade6"=c(9,6,9,6,7,6,9,9),
"Total_A_Grade7"=c(20,19,18,25,16,21,19,26),
"Group1_A_Grade7"=c(9,7,7,9,7,7,5,8),
"Group2_A_Grade7"=c(8,5,7,9,6,5,5,9),
"Total_B_Grade7"=c(25,21,24,25,18,18,27,18),
"Group1_B_Grade7"=c(10,10,10,7,5,6,8,5),
"Group2_B_Grade7"=c(9,6,8,10,8,6,10,6))
data2=data.frame("School"=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1),
"Fund"=c(0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1),
"Type"=c('Total','Total','Group1','Group1','Group2','Group2','Total','Total','Group1','Group1','Group2','Group2','Total','Total','Group1','Group1','Group2','Group2','Total','Total','Group1','Group1','Group2','Group2'),
"Class"=c('A','A','A','A','A','A','B','B','B','B','B','B','A','A','A','A','A','A','B','B','B','B','B','B'),
"Grade"=c(5,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6),
"Score"=c(22,20,10,6,5,9,23,33,8,7,6,10,18,24,7,7,5,8,26,23,10,10,9,6))
我有 'data1' 并且想要重塑以制作 'data2'，它只是展示了 School 1 5 年级和 6 年级的示例，但我想要重塑所有 data1。
'data1'的列名包含丰富的信息。例如，Group2_B_Grade6 表示 'Type' = Group2，'Class' = B，'Grade' = 6。我希望重塑 'data1'，然后使用这些由“分隔的存根” _" 作为 colnames 准备 'data2'
data3=data.frame("School"=c(1,1,2,2,3,3,4,4),
"Fund"=c(0,1,0,1,0,1,0,1),
"Grade_5"=c(22,20,21,24,24,26,25,22),
"Grade_6"=c(10,6,6,10,9,9,9,10),
"Grade_7"=c(5,9,9,8,10,8,8,6))
使用dplyr（和tidyr）：
library(dplyr)
library(tidyr)
data2 <- data1 %>%
pivot_longer(-c(School, Fund)) %>%
separate(name, into = c('Type', 'Class', 'Grade')) %>%
extract(Grade, 'Grade', "([0-9]+)")
data2
#> # A tibble: 144 x 6
#> School Fund Type Class Grade value
#> <dbl> <dbl> <chr> <chr> <chr> <dbl>
#> 1 1 0 Total A 5 22
#> 2 1 0 Group1 A 5 10
#> 3 1 0 Group2 A 5 5
#> 4 1 0 Total B 5 23
#> 5 1 0 Group1 B 5 8
#> 6 1 0 Group2 B 5 6
#> 7 1 0 Total A 6 18
#> 8 1 0 Group1 A 6 7
#> 9 1 0 Group2 A 6 5
#> 10 1 0 Total B 6 26
#> # … with 134 more rows
由 reprex package (v0.3.0)
于 2020-04-06 创建您可以直接使用 pivot_longer 和 names_pattern 中的一些正则表达式来执行此操作。
tidyr::pivot_longer(data1,
cols = -c(School, Fund),
names_to = c('Type', 'Class', 'Grade'),
names_pattern = '(.*?)_([A-Z])_Grade(\d+)',
values_to = 'Score')
# A tibble: 144 x 6
# School Fund Type Class Grade Score
# <dbl> <dbl> <chr> <chr> <chr> <dbl>
# 1 1 0 Total A 5 22
# 2 1 0 Group1 A 5 10
# 3 1 0 Group2 A 5 5
# 4 1 0 Total B 5 23
# 5 1 0 Group1 B 5 8
# 6 1 0 Group2 B 5 6
# 7 1 0 Total A 6 18
# 8 1 0 Group1 A 6 7
# 9 1 0 Group2 A 6 5
#10 1 0 Total B 6 26
# … with 134 more rows
我们可以使用 melt 来自 data.table
library(data.table)
melt(setDT(data1), id.var = c('School', 'Fund'))[,
c('Type', 'Class', 'Grade') := tstrsplit(variable, "_")][,
Grade := sub('Grade', '', Grade)][, variable := NULL][]
# School Fund value Type Class Grade
# 1: 1 0 22 Total A 5
# 2: 1 1 20 Total A 5
# 3: 2 0 21 Total A 5
# 4: 2 1 24 Total A 5
# 5: 3 0 24 Total A 5
# ---
#140: 2 1 10 Group2 B 7
#141: 3 0 8 Group2 B 7
#142: 3 1 6 Group2 B 7
#143: 4 0 10 Group2 B 7
#144: 4 1 6 Group2 B 7